Let's solve the limit:
[tex]\[
\lim_{{x \to 7}} \left( \frac{x^2 + x - 56}{x^2 - 11x + 28} \right).
\][/tex]
First, we need to simplify the expression, if possible. Let's factor both the numerator and the denominator.
The numerator is [tex]\(x^2 + x - 56\)[/tex]. We look for two numbers that multiply to [tex]\(-56\)[/tex] and add to [tex]\(1\)[/tex]. Those numbers are [tex]\(8\)[/tex] and [tex]\(-7\)[/tex]. So, we can factor the numerator as:
[tex]\[
x^2 + x - 56 = (x + 8)(x - 7).
\][/tex]
Next, we factor the denominator [tex]\(x^2 - 11x + 28\)[/tex]. We look for two numbers that multiply to [tex]\(28\)[/tex] and add to [tex]\(-11\)[/tex]. Those numbers are [tex]\(-4\)[/tex] and [tex]\(-7\)[/tex]. So, we can factor the denominator as:
[tex]\[
x^2 - 11x + 28 = (x - 4)(x - 7).
\][/tex]
Substituting these factorizations back into the limit, we get:
[tex]\[
\lim_{{x \to 7}} \left( \frac{(x + 8)(x - 7)}{(x - 4)(x - 7)} \right).
\][/tex]
We notice that both the numerator and the denominator have a common factor of [tex]\( (x - 7) \)[/tex]. As long as [tex]\( x \neq 7 \)[/tex], we can cancel out this common factor:
[tex]\[
\frac{(x + 8)(x - 7)}{(x - 4)(x - 7)} = \frac{x + 8}{x - 4}, \quad \text{for} \; x \neq 7.
\][/tex]
Now the limit becomes:
[tex]\[
\lim_{{x \to 7}} \left( \frac{x + 8}{x - 4} \right).
\][/tex]
We can now directly substitute [tex]\( x = 7 \)[/tex] into the simplified expression:
[tex]\[
\frac{7 + 8}{7 - 4} = \frac{15}{3} = 5.
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{{x \to 7}} \left( \frac{x^2 + x - 56}{x^2 - 11x + 28} \right) = 5.
\][/tex]
